Two families of lines are given by ax + by + | vedclass

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(D) Given the equation of the family of lines $ax + by + c = 0$ and the condition $4a^2 + 9b^2 - c^2 - 12ab = 0$. Rearranging the condition: $4a^2 - 1

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a line passing through $(2, -3)$ and $(-2, 3)$

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(D) Given the equation of the family of lines $ax + by + c = 0$ and the condition $4a^2 + 9b^2 - c^2 - 12ab = 0$. Rearranging the condition: $4a^2 - 12ab + 9b^2 = c^2$,which is $(2a - 3b)^2 = c^2$. Taking the square root on both sides,we get $c = \pm(2a - 3b)$. Case $1$: $c = 2a - 3b$. Substituting this into the line equation: $ax + by + (2a - 3b) = 0 \implies a(x + 2) + b(y - 3) = 0$. This line passes through the fixed point $(-2, 3)$. Case $2$: $c = -(2a - 3b) = -2a + 3b$. Substituting this into the line equation: $ax + by + (-2a + 3b) = 0 \implies a(x - 2) + b(y + 3) = 0$. This line passes through the fixed point $(2, -3)$. Comparing the options,the line passing through $(2, -3)$ and $(-2, 3)$ satisfies the condition.

Two families of lines are given by $ax + by + c = 0$ and $4a^2 + 9b^2 - c^2 - 12ab = 0$. Then the line common to both the families is

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